How to find the solution to an inequality with addition - PSAT Math
Card 0 of 42
What values of x make the following statement true?
|x – 3| < 9
What values of x make the following statement true?
|x – 3| < 9
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Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve for 
.
Solve for
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Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
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3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
If x+1< 4 and y-2<-1 , then which of the following could be the value of 
?
If x+1< 4 and y-2<-1 , then which of the following could be the value of
Tap to see back →
To solve this problem, add the two equations together:
x+1<4
y-2<-1
x+1+y-2<4-1
x+y-1<3
x+y<4
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
To solve this problem, add the two equations together:
x+1<4
y-2<-1
x+1+y-2<4-1
x+y-1<3
x+y<4
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
If 
, which of the following could be a value of 
?
If
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In order to solve this inequality, you must isolate 
on one side of the equation.
Therefore, the only option that solves the inequality is 
.
In order to solve this inequality, you must isolate
Therefore, the only option that solves the inequality is
What values of 
make the statement 
true?
What values of
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First, solve the inequality 
:
Since we are dealing with absolute value, 
must also be true; therefore:
First, solve the inequality
Since we are dealing with absolute value,
What values of x make the following statement true?
|x – 3| < 9
What values of x make the following statement true?
|x – 3| < 9
Tap to see back →
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve for .
Solve for
Tap to see back →
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
Tap to see back →
3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
If x+1< 4 and y-2<-1 , then which of the following could be the value of ?
If x+1< 4 and y-2<-1 , then which of the following could be the value of
Tap to see back →
To solve this problem, add the two equations together:
x+1<4
y-2<-1
x+1+y-2<4-1
x+y-1<3
x+y<4
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
To solve this problem, add the two equations together:
x+1<4
y-2<-1
x+1+y-2<4-1
x+y-1<3
x+y<4
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
If , which of the following could be a value of ?
If
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In order to solve this inequality, you must isolate on one side of the equation.
Therefore, the only option that solves the inequality is .
In order to solve this inequality, you must isolate
Therefore, the only option that solves the inequality is
What values of make the statement true?
What values of
Tap to see back →
First, solve the inequality :
Since we are dealing with absolute value, must also be true; therefore:
First, solve the inequality
Since we are dealing with absolute value,
What values of x make the following statement true?
|x – 3| < 9
What values of x make the following statement true?
|x – 3| < 9
Tap to see back →
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve for .
Solve for
Tap to see back →
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
If –1 < w < 1, all of the following must also be greater than –1 and less than 1 EXCEPT for which choice?
Tap to see back →
3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
3_w_/2 will become greater than 1 as soon as w is greater than two thirds. It will likewise become less than –1 as soon as w is less than negative two thirds. All the other options always return values between –1 and 1.
If x+1< 4 and y-2<-1 , then which of the following could be the value of ?
If x+1< 4 and y-2<-1 , then which of the following could be the value of
Tap to see back →
To solve this problem, add the two equations together:
x+1<4
y-2<-1
x+1+y-2<4-1
x+y-1<3
x+y<4
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
To solve this problem, add the two equations together:
x+1<4
y-2<-1
x+1+y-2<4-1
x+y-1<3
x+y<4
The only answer choice that satisfies this equation is 0, because 0 is less than 4.
If , which of the following could be a value of ?
If
Tap to see back →
In order to solve this inequality, you must isolate on one side of the equation.
Therefore, the only option that solves the inequality is .
In order to solve this inequality, you must isolate
Therefore, the only option that solves the inequality is
What values of make the statement true?
What values of
Tap to see back →
First, solve the inequality :
Since we are dealing with absolute value, must also be true; therefore:
First, solve the inequality
Since we are dealing with absolute value,
What values of x make the following statement true?
|x – 3| < 9
What values of x make the following statement true?
|x – 3| < 9
Tap to see back →
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve the inequality by adding 3 to both sides to get x < 12. Since it is absolute value, x – 3 > –9 must also be solved by adding 3 to both sides so: x > –6 so combined.
Solve for .
Solve for
Tap to see back →
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.
Absolute value problems always have two sides: one positive and one negative.
First, take the problem as is and drop the absolute value signs for the positive side: z – 3 ≥ 5. When the original inequality is multiplied by –1 we get z – 3 ≤ –5.
Solve each inequality separately to get z ≤ –2 or z ≥ 8 (the inequality sign flips when multiplying or dividing by a negative number).
We can verify the solution by substituting in 0 for z to see if we get a true or false statement. Since –3 ≥ 5 is always false we know we want the two outside inequalities, rather than their intersection.